# Simple projectile

1. Feb 18, 2010

### Sirsh

A long jumper is able to take off with a velocity of 8.25m/s at 12.5dgs to the horizontal. Calculate:

a) the initial horizontal velocity
a) 8.25m/s

b) the vertical take off velocity
b) 8.25sin(12.5) = 1.785m/s, as far as i know Uxsin(theta) is the vertical velocity , except in the answer to the second half of the question it says it's 8.25cos(12.5) which is the equation to the horizontal velocity? anyone know why this is? thanks.

2. Feb 18, 2010

### tiny-tim

Hi Sirsh!

(have a theta: θ and a degree: º )
Well, it's definitely sin12.5º for the vertical component.

The book must be wrong.

3. Feb 18, 2010

### Sirsh

Hey tiny-tim, thanks for the signs :P does the verical component be8ing 1.785m/s mean that after one second the jumper will be 1.785m in the air ect until he hits max parabolic height then he'll decend until he hits the horizontal plain? thanks (:

4. Feb 18, 2010

### tiny-tim

Hey Sirsh!

It means he'll always be at the same height as if he'd jumped vertically with that speed.

But it won't be 1.785m after 1 second, it'll be 1/2 at2.

5. Feb 18, 2010

### Sirsh

Oh okay, thanks alot! :D